Đáp án: $\left( x;y \right)=\left( 0;0 \right)\,;\,\left( 14;2 \right)$
Giải thích:
$\begin{cases}288x=21\left( {{x}^{2}}-{{y}^{2}} \right)\,\,\,\,\,\,\,\,\left(1\right)\\252xy-36{{y}^{2}}=36\left( {{x}^{2}}-{{y}^{2}} \right)\,\,\,\,\,\,\,\,\left(2\right)\end{cases}$
$\left( 2 \right)\Leftrightarrow 252xy-36{{y}^{2}}=36{{x}^{2}}-36{{y}^{2}}$
$\,\,\,\,\,\,\,\Leftrightarrow 36{{x}^{2}}-252xy=0$
$\,\,\,\,\,\,\,\Leftrightarrow {{x}^{2}}-7xy=0$
$\,\,\,\,\,\,\,\Leftrightarrow x\left( x-7y \right)=0$
$\,\,\,\,\,\,\,\Leftrightarrow\left[\begin{array}{1}x=0\\x=7y\end{array}\right.$
Khi $x=0$, thế vào phương trình $\left( 1 \right)$
Ta tìm được $y=0$
Khi $x=7y$, thế vào phương trình $\left( 1 \right)$
$\,\,\,\,\,\,\,288\,.\,7y=21\left[ {{\left( 7y \right)}^{2}}-{{y}^{2}} \right]=0$
$\Leftrightarrow 1008{{y}^{2}}-2016y=0$
$\Leftrightarrow\left[\begin{array}{1}y=0\\y=2\end{array}\right.$
$\to\left[\begin{array}{1}x=0\\x=14\end{array}\right.$
Vậy $\left( x;y \right)=\left( 0\,;\,0 \right)\,;\,\left( 14;2 \right)$
